Find the equation of the circle whose diameters are along lines $2x - 3y + 12 = 0$ and $x + 4y - 5 = 0$ and whose area is 154 sq. units.

Hint: To find the center of the circle we will find the intersection point of the given lines. After that, we will equate area 154 with $\pi {r^2}$ to calculate the radius. Then by using the standard form of the equation of a circle, we will find the equation of the circle.

Formula Used:
The standard formula of a circle is ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$, where $\left( {a,b} \right)$ is the center and $r$ is radius.
The area of a circle is $\pi {r^2}$ where $r$ is radius.

Complete step by step solution:
Given equations of lines are
$2x - 3y + 12 = 0$ …(i)
$x + 4y - 5 = 0$ ….(ii)
Multiply 2 with equation (ii) and subtract it from equation (i)
$\begin{array}{*{20}{c}}{2x}& - &{3y}& + &{12}& = &0\\{2x}& + &{8y}& - &{10}& = &0\\{\left( - \right)}&{\left( - \right)}&{}&{\left( + \right)}&{}&{}&{}\end{array}$
    $\overline {\begin{array}{*{20}{c}}{}& - &{11y}& + &{22}& = &0\end{array}} $
$ \Rightarrow 11y = 22$
Divide both sides by 11.
$ \Rightarrow y = 2$
Substitute the value of $y$ in equation (i)
$2x - 3 \cdot 2 + 12 = 0$
$ \Rightarrow 2x - 6 + 12 = 0$
$ \Rightarrow 2x + 6 = 0$
Subtract 6 from both sides
$ \Rightarrow 2x + 6 - 6 = - 6$
$ \Rightarrow 2x = - 6$
Divide both sides by 2
$ \Rightarrow x = - 3$
So, the center of the circle is $\left( { - 3,2} \right)$.
Assume that the radius of the circle is $r$.
The area of the circle is $\pi {r^2}$.
Equate $\pi {r^2}$ with 154
$\pi {r^2} = 154$
Put the value of $\pi = \dfrac{{22}}{7}$
$ \Rightarrow \dfrac{{22}}{7}{r^2} = 154$
Multiply both sides by $\dfrac{7}{{22}}$
$ \Rightarrow \dfrac{{22}}{7} \times \dfrac{7}{{22}}{r^2} = 154 \times \dfrac{7}{{22}}$
$ \Rightarrow {r^2} = 154 \times \dfrac{7}{{22}}$
Divide the denominator by 22 and numerator by 22
$ \Rightarrow {r^2} = 7 \times 7$
Take square root both sides
$ \Rightarrow r = 7$
Use the standard form of equation of circle to find the equation circle.
Here $h = - 3$, $k = 2$ and $r = 7$
${\left( {x - \left( { - 3} \right)} \right)^2} + {\left( {y - 2} \right)^2} = {7^2}$
$ \Rightarrow {\left( {x + 3} \right)^2} + {\left( {y - 2} \right)^2} = {7^2}$
Apply the algebraical identity on the left side terms
$ \Rightarrow {x^2} + 6x + 9 + {y^2} - 4y + 4 = 49$
$ \Rightarrow {x^2} + {y^2} + 6x - 4y + 9 + 4 - 49 = 0$
$ \Rightarrow {x^2} + {y^2} + 6x - 4y - 36 = 0$

Option ‘D’ is correct

Note: The question says that two diameters of the circle lie on the lines $2x - 3y + 12 = 0$ and $x + 4y - 5 = 0$. The intersection point of diameter is the center of a circle. To find an equation of a circle, we need the radius and center of the circle. We get the center from the intersection of the given lines. To find the radius, we use information about the area of the circle.